Question

Hi can you please help i cant seem to find an answer

Answer 1

Given data:

* The value of spring constant is,

`k=320Nm^(-1)`

* The displacement of the block is 0.5 m.

* The mass of the block is 0.6 kg.

Solution:

(a). The potential energy of the block is,

`U=(1)/(2)kx^2`

where k is the spring constant, and x is the displacement,

Substituting the known values,

`\begin{gathered} U=(1)/(2)*320*(0.5)^2 \\ U=40\text{ J} \end{gathered}`

The maximum kinetic energy of the block is,

`K=(1)/(2)mv^2`

where m is the mass of block and v is the maximum velocity,

By the law of conservation of energy,

`\begin{gathered} K=40 \\ (1)/(2)mv^2=40 \\ v^2=(40*2)/(m) \\ v^2=(80)/(0.6) \\ v^2=133.33 \\ v=11.55\text{ m/s} \end{gathered}`

The maximum velocity of the block is 11.55 m/s.

(b). The velocity of the block at the distance 0.16 m is,

`\begin{gathered} (1)/(2)mv^2=(1)/(2)kx^2 \\ mv^2=kx^2 \\ v^2=(k)/(m)x^2 \end{gathered}`

Substituting the known values,

`\begin{gathered} v^2=(320)/(0.6)*0.16^2 \\ v^2=13.65 \\ v=3.7\text{ m/s} \end{gathered}`

Thus, the velocity of the box at 0.16 m is 3.7 m/s.

(c). The force acting on the box is,

`F=kx`

Substituting the known values,

`\begin{gathered} F=320*0.16 \\ F=51.2\text{ N} \end{gathered}`

By the Newton's second law,

The acceleration of the box is,

`\begin{gathered} F=ma \\ a=(F)/(m) \end{gathered}`

Substituting the known values,

`\begin{gathered} a=(51.2)/(0.6) \\ a=85.33ms^(-2) \end{gathered}`

Thus, the acceleration of the box is 85.33 meter per second squared.

(d). The restorign force acting on the box is opposite to the direction of motion of the box.

Thus, the restoring force acting on the box is -51.2 N.